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If I have that $X_n$ is a two-state Markov chain whose transition probability matrix is:

$P = \left( \begin{smallmatrix} \alpha & 1-\alpha\\ 1-\beta & \beta \\\end{smallmatrix} \right)$

Then $Z_n=(X_{n-1},X_n)$ is a Markov chain having the four states (0,0), (0,1), (1,0) and (1,1). How would I determine the transition probability matrix? I appreciate all of the help and suggestions!

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the possible transistions are those for which $Z_{n,2} = Z_{n+1, 1}$, ie $$ (0,0) \to (0, 0)\\ (0,0) \to (0, 1)\\ (1,0) \to (0, 0)\\ (1,0) \to (0, 1)\\ (0,1) \to (1, 0)\\ (0,1) \to (1, 1)\\ (1,1) \to (1, 0)\\ (1,1) \to (1, 1) $$

and for instance the transition $(0,1)\to (1,1)$ has probability $$ P(X_{n+1} = 1, X_{n} = 1 | X_{n} = 1, X_{n-1} = 0) = P(X_{n+1} = 1, X_{n} = 1 | X_{n} = 1) $$ because of the Markov property, $$ = P(X_{n+1} = 1| X_{n} = 1) = \beta $$

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  • $\begingroup$ so would the matrix be in terms of numbers or in terms of $\alpha$ and $\beta$? $\endgroup$
    – user141745
    Feb 3 '15 at 19:29
  • $\begingroup$ can't you compute it with the help of my message? $\endgroup$
    – mookid
    Feb 4 '15 at 23:05
  • $\begingroup$ @user141745 the matrix would be in terms of $\alpha$, $1-\alpha$, $\beta$, and $1-\beta$ since that is what was given in the two-state Markov matrix. If it's helpful to think about it that way then consider these terms as probabilities such that $\sum_j p_{ij}=1$. $\endgroup$
    – user711703
    May 24 '20 at 16:50

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